The fractional part of a real number x is def | vedclass

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Question

(C) Let $x_1 = (44 + \sqrt{2017})^{2017}$ and $x_2 = (44 - \sqrt{2017})^{2017}$.Since $44^2 = 1936$ and $45^2 = 2025$,we have $44 < \sqrt{2017} < 45$.

Vedclass · Accepted Answer

$0.9$ and $1.35$

Vedclass · Answer

(C) Let $x_1 = (44 + \sqrt{2017})^{2017}$ and $x_2 = (44 - \sqrt{2017})^{2017}$.Since $44^2 = 1936$ and $45^2 = 2025$,we have $44 Let $x_1 = I + F_2$,where $I$ is an integer and $0 \le F_2 Since $0 Consider $x_1 + x_2 = (44 + \sqrt{2017})^{2017} + (44 - \sqrt{2017})^{2017}$.By the binomial expansion,the irrational terms involving $\sqrt{2017}$ cancel out,leaving $x_1 + x_2 = 2 \sum_{k=0, 	ext{even}}^{2017} \binom{2017}{k} 44^{2017-k} (2017)^{k/2}$,which is an even integer $N$.Thus,$I + F_2 + F_1 = N$,which implies $F_1 + F_2 = N - I$. Since $0 The only integer value $F_1 + F_2$ can take is $1$.Since $1$ lies between $0.9$ and $1.35$,the correct option is $C$.

The fractional part of a real number $x$ is defined as $x - [x]$,where $[x]$ is the greatest integer less than or equal to $x$. Let $F_1$ and $F_2$ be the fractional parts of $(44 - \sqrt{2017})^{2017}$ and $(44 + \sqrt{2017})^{2017}$,respectively. Then,$F_1 + F_2$ lies between the numbers:

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